You can use https://yandex.ru/cup/profile.

When we can upsolve/discuss the problems?

I did not solve this problem in contest but here is what i think is the correct solution.

First of all, let's try to reason about given constraints in terms of prefix sums. You also should note that since any song can have any mood, any prefix sum array can be constructed.

Let a be the array of the song moods.

Then sum(a[l..r]) <= x is equivalent to p[r] — p[l — 1] <= x, where p[i] is the sum(a[1..r]).

Also consider that you can rewrite ">=" constraint in terms of "<=", so:

p[l — 1] — p[r] <= -x

Okay, what's next? It looks like directed weighted graph problem now, where there's either an edge (l-1; ? with weight x ("<=" case), or edge (r; l-1) with weight -x (">=" case). But how does that help?

You should notice that we can even think about putting a "No" as the answer only when there's a cycle in the graph. Moreover, if you see that this cycle is of negative weight, this will lead to a "No" answer, since in the end some p[i] has to be smaller than you started with every time you go through the negative cycle.

So the problem is reduced to find if there's a negative weight cycle, which can be done with Bellman-Ford algorithm in $$$O(NM)$$$

В прошлом году одна D гарантировала полуфинал(или тогда сразу финал был), возможно не будет кардинальной разницы

I've got a checkmark on a mobile track, but not algorithms. What's interesting is that you can only choose one track for semi-finals. And the mobile track semi-finals has already started. So while I'm waiting for the results to make a decision — I'm wasting my time if I stick to a mobile track.

I have asked about the criteria for qualification during contest, and here is their response: "Hello! We will publish and send out the criteria for the next stage after the end of the contest within two or three days."

UPD:got a checkmark.

Yeah would have been better if they would have told about percentage or number of people to qualify. I surely thought qualification is basic check and ABC should be enough

I advanced with ABCE. Looks like the criteria is somewhere between 3 and 4 problems. Looks like all problems have equal weights since the score is given as number of problems solved.